Lesson video

Thank you for choosing to learn using this video today.

My name is Ms. Davis and I'm gonna help you as you work your way through this lesson.

Feel free to pause the video and work through things at your own pace.

Lots of exciting things to look at.

So let's get started.

Welcome to today's lesson.

Our lesson title is Many Equations with One Solution.

Today's lesson outcome, "I can understand that a family of linear equations can all have the same solution." Keyword that we're gonna use a lot today is solutions.

The solutions to an equality are the values which when substituted maintain the equality between the expressions.

We're gonna start by looking at maintaining equality while manipulating equations.

"If we know two expressions are equal, we can form an equation.

If we perform the same operation to both expressions, they remain equal to each other." Let's look at an example.

We've got 8 + 4 = 12.

If I add 1 to the expression on the left-hand side, it no longer equals 12.

8 + 4 + 1, which is now 13.

What would you need to do to the right-hand side to maintain equality? Well done.

We need to add 1 to 12 as well.

These expressions have now both changed, but they are now equal to each other.

8 + 4 + 1 = 12 + 1.

What would I need to do to maintain equality this time? I've added -10 to the left-hand side, I'll need to add -10 to the right-hand side.

If you want to write the right-hand side as a single term, 12 add -10 is 2.

We can check that we have maintained equality by seeing if both sides of the equation are still equal.

8 + 4 - 10 is the same as 12 - 10, which gives us 2.

So I've got 2 = 2.

So we have maintained equality.

The left and the right-hand side's still balance.

What will happen if I double the expression on the right-hand side? Well now I've got 12 x 2, which we could write as 24, if we wish.

What would we need to do to maintain equality? Think carefully about this one.

We need to multiply the whole of the left-hand side by 2.

I've got 8 + 4.

I need to multiply all of that by 2.

Notice that brackets are necessary to ensure that the whole expression is multiplied by 2.

And the convention is to write the multiplier at the front of the bracket.

Laura says, "To check, we could calculate 8 + 4 then multiply that by 2." Alex says, "We should do 2 x 8 and 2 x 4 and add them together." Do you agree with either statement? Let's check it out.

8 + 4 multiplied 2 = 24.

12 x 2 = 24.

24 = 24.

So we have maintained equality.

We've shown that with Laura's method.

Let's look at Alex's method.

If we do 2 multiplied 8 and 2 multiplied 4 and add them together, we get 16 + 8.

Just 24 = 24.

So again, we have shown that we have maintained equality.

Both methods will work.

Alex uses the distributive property to expand the bracket and Laura evaluates according to the priority of operations.

I'd like you to have a look at this question.

What mistake has been made? How could you then show that equality has not been maintained? What do you think? Okay, the mistake has been made when dividing the left-hand side by 4.

The whole expression should be divided by 4, not just the 4.

With priority of operations, if you just write divide by 4 at the end, that means you're only dividing the four by four.

You need to divide the whole thing by four.

We could check it by saying the left-hand side evaluates to 8 + 1, which is 9, the right-hand side evaluates to 3.

So we can't have maintained equality 'cause the left and the right are not the same anymore.

Let's look at what this should look like.

We can use brackets or even better, we can use fractional notation 'cause that means the whole of the numerator divided by 4.

We can do the same thing with equations that have variable terms. This is an equation where the solution is clearly when x is 5, x = 5.

We can create an equivalent equation by adding the same value to both sides.

If I add 3, I can write that as x + 3 = 5 + 3.

I could also write that as x + 3 = 8, if I wish.

I can add a negative value.

So this equation could become x add -7 = 5 add -7.

I've maintained equality because I've added the same value to both sides.

I can add a variable.

So I can add x.

So x add x = 5 add x.

I could also write that as 2x = x + 5 by collecting like terms. We can also create equivalent equations by multiplying both sides by the same value.

So if I multiply by 3, this would look like 3x = 15.

If I multiply by -4, this would be -4x = -20.

I could multiply by a fraction.

If I multiply by 1/3, I would get 1/3x = 5/3.

Multiplying by 1/3 is the same as divided by 3.

So you could have written this as x/3 = 5/3.

1/3x and x/3 three are equivalent.

Time to have a check.

I'd like you to complete the diagram.

You've got a box at the end of each row to fill in.

Then the middle row, you've got a missing step as well.

Give it a go.

Let's check our answers.

So you should have 2x = -36.

The middle one, we would like to write that as x + 9 = -18 + 9.

Then I can write that as x + 9 = -9.

On the right-hand side, if you want to divide by -6, so that's x/-6.

You could have also written that as -x/6 or -1/6x and that's gonna equal 3.

Remember, -18 divided by -6 is gonna be +3.

"Because the initial equation in the centre of our diagram has a solution when x is 5, all the equivalent equations will have a solution of x = 5." We can substitute the solution in each equation to check.

So let's look at 3x = 15.

Does that balance when x is 5 or 3 x 5 is 15? 15 = 15, that's balanced.

5 + 3 = 8.

8 = 8, that is balanced.

We can do the same for 2x.

X subtract 7 and -4x.

We can clearly see that when x is a solution to one of the equations, it's also a solution to all the equivalent equations.

We can perform multiple operations to an equation as long as we perform the same operation to the expressions on both sides, we will maintain equality.

So let's start with the simple equation, x = 8 and add 3, that gives us x + 3 = 11.

We could then multiply by 2, we get two lots of x + 3 = 22.

Using brackets ensures that the whole expression is multiplied by 2.

If we wish, we could expand the brackets, which gives us 2x + 6.

When expanding the brackets, no operation has been performed, the expression is just being written in a different format.

So the right-hand side is still 22.

That's equivalent to what's written above it.

Let's look at this example.

If I multiply by 2, I could write that as 2x = 16.

Then add 3, I can write that as 2x + 3 = 19.

I now want to multiply the whole thing by three, how would we maintain equality for this step? Lovely, use our brackets to make sure we multiply the whole expression.

This is the same as multiplying each term by 3.

So you could have written it as 6x + 9 = 57 instead.

What about if I just want to add 3? Add three.

Now I've bracketed it just to show you what's happening.

We're adding three to the whole of that left-hand side.

But adding three to the whole expression is the same as adding three to a constant term.

If you think about it with numbers, if I had 8 add 4 add 3, I could do 8 add 4 and then add 3 to it or that would be the same as doing 8 add 4 add 3.

So it doesn't matter where you write that add three, when we collect like terms, you add the constants together.

So that would be 2x + 6 = 22.

We can start with some more complex equations too.

Let's have a look at some.

Let's add 3x - 2 = 8.

If I add three this time, remember, I'm adding three to the constant term.

That's 3x + 1 = 11 and then multiplying by two.

Use a bracket, if I wish.

Let's see what happens when I add -x.

Now adding -x to the whole expression is like adding -x to the x10.

So 3x add -x is 2x and then we've got 8 subtract x or 8 add -x on that right-hand side.

Let's try and multiplying that by 1/2.

Using brackets, 1/2 of 2x - 2 = 1/2 of 8 - x.

Depending on what you then want to do with that, you might want to write that using division instead.

So we could write that as 2x - 2/2 = 8 - x/2 And then we could simplify each individual term.

So 2x divided by 2 is x, <v ->2 divided by 2 is 1,</v> 8 divided by 2 is 4 and -x divided by 2 is -x/2 or -1/2x.

Let's multiply by five.

So we'd have 5 lots of 3x - 2 = 40.

If I then wanted to add 2x, I can just put add 2x.

If I wanted to, I could expand the bracket and collect like terms, and there we go.

Time for you to have a go.

Given that 3x + 11 = 10x + 5, which of these are true? There's quite a few to read, so give yourself enough time to give it a good go.

Well done if you spotted that the top one, the second one and D are all true.

The first one, one has been added to both sides.

The second one, both sides have been doubled.

The D, x has been subtracted from both sides.

Time to have a practise.

I'd like you to complete the equations in this diagram.

The arrows show the operation that is being applied to the whole equation.

Give those a go and come back for the answers.

Well done, there's lots going on in that diagram.

So I'm gonna start the top left-hand row and then I'm gonna move clockwise.

So to start with, when we're adding four, x + 4 = 7.

Then we're multiplying by five, that's 5 lots of x + 4 = 35.

And then adding -4, 5 lots of x + 4 - 4 = 31.

If you chose to expand your bracket, you might have written that first step as 5x + 20 and then the second one would be 5x + 20 add -4 or 5x + 16 = 31.

All of those are equivalent.

Let's have a look at the top right-hand.

So adding 5x, we've now got 6x = 3 + 5x.

Adding 1, 6x + 1 = 4 add 5x.

Then adding -5x, we've got x + 1 = 4.

Bottom right-hand spike.

Multiplying by 1/3, x/3 = 1 or 1/3x = 1.

Then I'm gonna add x + 1.

So you've got 1/3x + x + 1 = x + 2 or x/3 + x + 1 = x + 2.

For the final one, multiplying by three, gives us 3x = 9.

Adding -10, gives us 3x - 10 = -1.

And dividing by 3, that's 3x = 10 all over 3 = -1/3.

Well done.

Let's look at the next part of our lesson.

In this next part of the lesson, we're gonna look at generating equations the same solution.

If equations are equivalent, we can work out the operations which link them.

So here is part of a spider diagram.

What operation goes on each arrow? Okay, so the first one, multiplying by -5.

Moving clockwise, dividing by -4 or multiplying by -1/4.

Then we've got adding seven, adding 2x, and adding x - 2.

Well, I dunno if you spotted that one.

It's the same as adding x and adding -2.

All the operations produce equivalent equations.

They've all changed in some ways, but they're equivalent to the original equation.

"Some operations will make the equation look more complicated and some may make it look simpler.

What would the equations in this diagram look like?" Take your time, write your answers for each one.

Top right, I've got 2x - 10 = 10, 2 lots of 2x - 6 = 28 or 4x - 12 = 28.

2x = 20, and -6 = 14 - 2x.

Top one, you could have written as 2x - 6 all over 2 = 7.

But all those terms, divide nicely by two, so I could have written that as x - 3 = 7.

Well done if you got those yourself.

Which operations seem to make the equation more complicated? Were there any operations that seemed to make it look simpler? What do you think? Okay, you may have said that adding -2x or multiplying by 2 potentially make it look a little bit trickier.

You may have said that divided by 2 or adding 6 seem to have made the equation look simpler.

Andeep says, "I think the solution might be when x is 10." Which equation do you think Andeep might have used to suggest this solution and how can we check? Well, all the equations will have the same solution as they're all equivalent.

However, I think that those two are the simplest.

So maybe x - 3 = 7 or maybe the 2x = 20.

We can check by substituting x = 10 into any of the equations.

We'll try them all.

10 - 3 = 7.

7 = 7.

Yes, that balances.

2 lots of 10 - 6 = 14.

20 - 6 = 14.

14 = 14.

That is correct.

And then 2 x 10 = 20.

20 = 20.

Yes, x = 10 is the solution to all of those.

This time, the missing number in the centre of the diagram is representing the solution to all these equations.

What operation goes on each arm this time? Okay, we've got multiply by 3, add 5.

This is a tricky one.

Multiply by -1 or divide by -1 and then add x + 1.

I'd like you to just take a moment.

Do you think you know what the solution to all these equations are? If you have an answer, hold onto it.

We are gonna try some values to check.

I'm picking this bottom right-hand equation to use 'cause I think it looks the easiest.

So x + 5 = 3.

Let's say x was 1.

That would give me 1 + 5 = 3, which gives me 6 = 3.

Well, that's not true.

They're not balanced.

Solution can't be x = 1.

Let's try x = 0.

0 + 5 = 3, 5 = 3.

Okay, still not true.

I need a smaller number.

Let's try -2.

<v ->2 + 5 = 3.

</v> 3 = 3.

That's looking likely.

So I think the solution is when x = -2.

Is that what you got first time around? We can actually check with the other equations that that works.

So let's use our x = -2 in the 3x = -6.

So 3 lots of -2 = -6.

<v ->6 = -6.

That is balance.

</v> Let's try the trickiest-looking one.

2 lots of -2 add 1 = -2 subtract 1.

So that's -4 add 1 = -3, and then that is balance.

So we know that our solution is when x = -2.

Okay, given that 20 = 10 subtract 2x, which of these are true? Take your time, come back for the answers.

Lovely.

A, D, and E this time.

2x has been added to both sides in A.

In B, 1 has been added to the left, but x + 1 has been added to the right.

C, we could check that solution by substituting.

It does not work.

D, 10 has been subtracted from both sides.

And E, 20 has been added to both sides.

Time to put all those skills into practise.

So very similar diagram to task A.

However, this time, some of the operations are missing as well.

So fill in the missing expressions to make all the equations equivalent and fill in the missing operations.

Don't forget the simple equation in the middle that shows us the solution.

Take your time, come back for the answers.

Well done.

You might have done these in different orders.

Just check that you've got the same answers as me.

I'll leave it up on the screen at the end.

Make sure you check.

So I think possibly the easier one to start with was the right-hand one top right.

So I'm gonna do that one first.

So we're adding 10.

So the left-hand side is gonna read x + 10.

Then to get from 6 to 2, we're adding -4.

So that's gonna go on our operation.

And then our final one, we're adding -x, so that gives us 6 = 2 - x or 2 add -x.

You could have used that one to work backwards to the value in the middle.

So the value in the middle would then be -4.

Okay, let's do the bottom right-hand one.

So I'm multiplying by 1/3, so I can write that as 1/3x or x/3 on the left-hand side.

If we didn't know what the right-hand side was, we'd have to use that bottom right one.

However, because we've worked out the middle is x = -4, you could write that as -4/3.

And then when we add x, that does give us what we need in the bottom right-hand corner.

So looking at that bottom right-hand one, we've got the x/3 + x = -4/3 + x or you could write that as x - 4/3, which is the way I've written it.

Okay, let's do the bottom left-hand one and then we'll come up to the top left-hand one.

So the bottom left-hand one, we're adding 3x.

So that should be 4x = 3x - 4.

We could then use that to work out.

The middle should be x = -4, if we hadn't already.

Then our missing operation is adding 4x.

Be careful that you didn't put double.

The left-hand side has doubled, but the right-hand side hasn't.

So it's add 4x to both sides.

And then to add 4, we can write that as 8x + 4.

And then if I've got -4 and I add 4, there's zero pairs.

So I'll be left with just 7x on the right-hand side.

And then the top left to finish, I'm multiplying by 2, so that'll give me 2x.

If you'd already worked out that the middle was -4, you know that's gonna be -8.

Then to get to the 1, you need to add 9.

So that gives you 2x + 9 = 1.

And then we already know that the last one is 2x + 9/5 = 1/5 If you hadn't worked out that the middle was -4, you could have used that final bit.

The 2x + 9/5 = 1/5 to work out the previous bit because you could have seen that to get from 1 to 1/5, you multiply by 1/5.

And then 2x + 9 multiply by 1/5 gives you 2x + 9/5.

Well done, do use this time to check you've got all the correct answers and then I'll see you to bring our lesson to a close.

Well done today.

So we have learned that given a single value of a variable, we can write equivalent equations by applying the same operation to expressions on both sides.

We know that the solution to all the equivalent equations will be the same.

We've talked about how some operations make the equation look more complicated and some may make it look simpler.

It is possible to perform operations which make the solution to the equation clear.

Thank you for joining me today and I really hope you choose to learn with us again.